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Unformatted text preview: 1 7. Dynamical System 1order Firstorder system: The differential equation is Solution: Take Laplace Transform: If u(t) is a unit step, ) t ( u ) t ( y dt ) t ( dy a = + 1 as ) s ( U ay ) s ( Y ) s ( U ) s ( Y ay ) s ( asY o o + + = = + 1 as ay ) s ( Y s 1 o + + = 2 1order systems (cont.) Let y o = 0, i.e. no initial condition then {} a t a 1 e 1 s 1 L L ) t ( y s 1 s 1 Y(s) a s 1 lim B , 1 1 as 1 lim A 1 as B s A ) 1 as ( s 1 ) s ( Y a 1 1 s 1 1 a 1 s s  = + = + =  = = = + = + + = + = 62.3% 1 y(t) t 3 1order systems time constant The timeconstant is defined as the time taken to reach the full value if the initial rate is maintained. Initial rate is Therefore the time to reach 1 takes a seconds; and at t = a And the time constant is a seconds. t dt dy = 63.2% i.e. , 632 . e 1 ) t ( y 1 = = a 1e1 1 y(t) t ( 29 a 1 t e e 1 dt d dt dy a t a t a 1 = = = = 4 Dynamical Systems Behavior (2order) We look at a singledegreeof freedom (sdof) system first. Usually we represent this system by a springmass damper system as shown. m c k x(t) f(t) The reason we call it sdof is: only one variable, i.e....
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This note was uploaded on 11/09/2009 for the course ENG 91301 taught by Professor Lui during the Spring '08 term at Hong Kong Institute of Vocational Education.
 Spring '08
 LUI
 Laplace

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